Ta có:
\(x=\dfrac{2006}{2007}-\dfrac{2007}{2008}+\dfrac{2008}{2009}-\dfrac{2009}{2010}\)

\(=\dfrac{2006.2008-2007^2}{2007.2008}+\dfrac{2008.2010-2009^2}{2009.2010}\)

\(=\dfrac{2006.2007+2006-2007^2}{2007.2008}+\dfrac{2008.2009+2008-2009^2}{2009.2010}\)

\(=\dfrac{2007\left(2006-2007\right)+2006}{2007.2008}+\dfrac{2009\left(2008-2009\right)+2008}{2009.2010}\)

\(=\dfrac{-1}{2007.2008}+\dfrac{-1}{2008.2010}< \dfrac{-1}{2006.2007}+\dfrac{1}{2007.2008}\)

\(\Rightarrow x< y\)

Vậy x < y

1/x+x+1+x+2+x+3+...+x+2006+2007=2007

------------------------------------------=2007-2007

------------------------------------------=0

x+x+x+...+x+1+2+3+...+2006=0

2007.x+(1+2+...+2006)=0

2007.x+(2006+1).[(2006-1)+1]:2=0

2007.x+2013021=0

2007.x=0-2013021

x=-2013021:2007

x=-1003

2/x+x+1+x+2+...+x+198=401-201-200-199

199.x+(1+2+...+198)=-199

199.x+(1+198).[(198-1)+1]:2=-199

199.x+19701=-199

199.x=-199-19701

x=-19900:199

x=-100

3/x+x+1+x+2+...+x+2008=2010-2010-2009

2009.x+(2008+1).[(2008-1)+1]:2=-2009

2009.x+2017036=-2009

2009.x=-2009-2017036

x=-2019045:2009

x=-1005

\(\frac{x-1}{2009}+\frac{x-2}{2008}=\frac{x-3}{2007}+\frac{x-4}{2006}\)

\(\Leftrightarrow\frac{x-1}{2009}-1+\frac{x-2}{2008}-1=\frac{x-3}{2007}-1+\frac{x-4}{2006}-1\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}\ne0\right)=0\)

\(\Leftrightarrow x=2010\)

!x-2007!+!x-2010!>=3 đẳng thức khi 2007<=x<=2008 

!x-2007!+!x-2008!+!x-2010!>=3 đẳng thức khi !x-2008!=0 

=> nghiệm duy nhất x=2008 và y=2009

\(\frac{x+1}{2010}+\frac{x+2}{2009}+\frac{x+3}{2008}=\frac{x+4}{2007}+\frac{x+5}{2006}+\frac{x+6}{2005}\)

<=> \(\frac{x+1}{2010}+1+\frac{x+2}{2009}+1+\frac{x+3}{2008}+1=\frac{x+4}{2007}+1+\frac{x+5}{2006}+1+\frac{x+6}{2005}+1\)

<=> \(\frac{x+2011}{2010}+\frac{x+2011}{2009}+\frac{x+2011}{2008}-\frac{x+2011}{2007}-\frac{x+2011}{2006}-\frac{x+2011}{2005}\) =0

<=> (x+2011).(\(\frac{1}{2010}+\frac{1}{2009}+\frac{1}{2008}-\frac{1}{2007}-\frac{1}{2006}-\frac{1}{2005}\) )=0

<=> x+2011=0

<=> x=-2011

Vậy pt có nghiệm là x=-2011

\(\Leftrightarrow\left(\frac{x-1}{2009}-1\right)+\left(\frac{x-2}{2008}-1\right)=\left(\frac{x-3}{2007}-1\right)+\left(\frac{x-4}{2006}-1\right)\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}=\frac{x-2010}{2007}+\frac{x-2010}{2006}\)

\(\Leftrightarrow\frac{x-2010}{2009}+\frac{x-2010}{2008}-\frac{x-2010}{2007}-\frac{x-2010}{2006}=0\)

\(\Leftrightarrow\left(x-2010\right)\left(\frac{1}{2009}+\frac{1}{2008}+\frac{1}{2007}+\frac{1}{2006}\right)=0\)

\(\Rightarrow x-2010=0\Rightarrow x=2010\)

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